---
title: "Reciprocal system of vectors"
subject: "Mathematics"
chapter: 5
key: "semester_2/mathematics:reciprocal-system-vectors"
source: https://www.bctengineer.com/q/semester_2/mathematics/reciprocal-system-vectors
site: BCT Engineering
---

# Reciprocal system of vectors

Mathematics · Chapter 5: Vector Algebra and Calculus

- Asked in: 2074 Bhadra, 5 marks; 2073 Magh, 5 marks; 2071 Magh, 5 marks (asked 3 times)
- Video: [Vectors  - ( Reciprocal system of vectors ; Solving problems ) - 84.](https://www.youtube.com/watch?v=MHjF5GkihnU) — Ameen Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/reciprocal-system-vectors
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-5

## Question

Find the set of reciprocal system to the set of vectors 2i+3j−k, −i+2j−3k and 3i−4j+2k.

## Short answer

Calculate [a b c] = -35. Reciprocal vectors: a' = (b×c)/[a b c] = (8i+7j+2k)/35, b' = (c×a)/[a b c] = (2i -7j -17k)/35, c' = (a×b)/[a b c] = (i - j - k)/5.

## Answer

For vectors a, b, c with [a b c] = a · (b × c) ≠ 0, the reciprocal system is

a′ = (b × c)/[a b c],   b′ = (c × a)/[a b c],   c′ = (a × b)/[a b c],

so that a · a′ = b · b′ = c · c′ = 1 and a · b′ = a · c′ = … = 0.

Here a = (2, 3, −1), b = (−1, 2, −3), c = (3, −4, 2).

#### Cross products
```text
b × c = (2·2 − (−3)(−4), (−3)(3) − (−1)(2), (−1)(−4) − 2·3)
      = (4 − 12, −9 + 2, 4 − 6) = (−8, −7, −2)
c × a = ((−4)(−1) − 2·3, 2·2 − 3(−1), 3·3 − (−4)(2))
      = (4 − 6, 4 + 3, 9 + 8) = (−2, 7, 17)
a × b = (3(−3) − (−1)(2), (−1)(−1) − 2(−3), 2·2 − 3(−1))
      = (−9 + 2, 1 + 6, 4 + 3) = (−7, 7, 7)
```

#### Scalar triple product
[a b c] = a · (b × c) = 2(−8) + 3(−7) + (−1)(−2) = −16 − 21 + 2 = −35.

#### Reciprocal vectors
```text
a′ = (−8i − 7j − 2k)/(−35) = (8i + 7j + 2k)/35
b′ = (−2i + 7j + 17k)/(−35) = (2i − 7j − 17k)/35
c′ = (−7i + 7j + 7k)/(−35) = (i − j − k)/5
```

Check: a · a′ = (16 + 21 − 2)/35 = 1 ✓; a · b′ = (4 − 21 + 17)/35 = 0 ✓; b · c′ = (−1 − 2 + 3)/5 = 0 ✓.

**Answer: a′ = (8i + 7j + 2k)/35, b′ = (2i − 7j − 17k)/35, c′ = (i − j − k)/5.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The reciprocal vector a' is:
   A. (a × b)/[a b c]
   B. (b × c)/[a b c]
   C. (b × c)·a
   D. (b × c)/|a|
   Answer: B. a' = (b×c)/[abc], b' = (c×a)/[abc], c' = (a×b)/[abc]. (Key point: Formula)

2. For a = (2,3,−1), b = (−1,2,−3), c = (3,−4,2), [a b c] equals:
   A. −35
   B. 35
   C. −16
   D. −21
   Answer: A. a · (−8,−7,−2) = −16 − 21 + 2 = −35. (Key point: Triple product)

3. a' equals:
   A. (−8i − 7j − 2k)/35
   B. (8i + 7j + 2k)/(−35)
   C. (2i − 7j − 17k)/35
   D. (8i + 7j + 2k)/35
   Answer: D. Dividing (−8,−7,−2) by −35 flips the sign. (Key point: Result)

4. The reciprocal system satisfies:
   A. a · a' = 0 and a · b' = 1
   B. a × a' = 1
   C. a · a' = 1 and a · b' = 0
   D. a' = a
   Answer: C. Dot products with the same-index vector are 1, others 0. (Key point: Defining property)

## Close variants set in other exams

- [Reciprocal system of 2i−3j+k, i+2j−k, 3i−j+2k](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Areciprocal-system-2i-3j-k) — Mathematics, chapter 5, asked 4×

## Related in Mathematics

- [Reciprocal system of 2i+3j−k, i−j−2k, −i+2j+2k](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Areciprocal-system-2i-3j-k-2) — Mathematics, chapter 5, asked 2×